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My first idea were some factorization-based solution. For example, adding 1 to both sides, and then:

$$n^3-n-1=k^2-k+1$$

$$n^3-n=k^2-k+2$$

$$(n-1)n(n+1)=k^2-k+2$$

...but I don't have idea, what to do with the right side. Any hint?

peterh
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  • Note that the left hand side is divisible by $3$... – Myself Dec 18 '14 at 21:38
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    Just as a matter of perspective, you are looking for integral points on the elliptic curve $y^2 - y = x^3 - x - 2$. If the elementary number theory doesn't pan out, we might need to look at slightly deeper techniques. – hardmath Dec 18 '14 at 21:42

1 Answers1

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Hint: Note that: $n^3 - n \equiv 0 \pmod 3$, thus

$4(k^2-k+1) = 3(k-1)^2 + (k+1)^2 \equiv 4(n^3-n-1) \equiv -1 \pmod 3$

But, $(k+1)^2 \equiv 0,1 \pmod 3$

r9m
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